A 41 kg block is held at rest on a frictionless ramp of angl

A 4.1 kg block is held at rest on a frictionless ramp of angle 56°. A bullet of mass 21 g shot up and parallel to the ramp, into the middle of the block. Before entering the block, it has a speed of 1084 m/s. Upon leaving the block through the other side it has a speed of 242 m/s. The block then moves up the ramp, unhindered from any upward motion by what was originally holding it in place. How far up the ramp in meters does the block go?

the kinetic energy lost by the bullet is equal to the difference of kinetic energy before and after entering the block

intial kinetic energy K1 = mv1^2/2 and ; v1 = 1084 m/s ; m = 21gm or 0.021kg

final kinetic energy = K2 = mv2^2 / 2 ; v2 = 242 m/s and m = 21gm or 0.021kg

difference K1-K2 = 11723.166 J

the difference in the kinetic energy is utilized by the 4.1 kg block to climb up the frictional less ramp

if we consider that the final postiion of the block at a distance S ; the corresponding height be h

sin(q) = h / S => h = S * sin(q)

the gain in the gravitational potential energy of the block by climbing a height h is equal to the kinetic energy lost by the bullet

M * g *h = K1-K2

4.1 kg * 9.8 m/s^2 * S * sin(56) = 11723.166 J

S = 11723.166 / 4.1 * 9.8 * sin(56)

S = 347 m is the distance moved by the block on the ramp

A 41 kg block is held at rest on a frictionless ramp of angl

Solution

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